Infinite Geometric Series

There is a simple test for determining whether a geometric series converges or diverges; if \(-1 < r < 1\), then the infinite series will converge. If \(r\) lies outside this interval, then the infinite series will diverge.

Test for convergence:

If \(-1 < r < 1\), then the infinite geometric series converges.
If \(r < -1 \text{ or } r > 1\), then the infinite geometric series diverges.

We derive the formula for calculating the value to which a geometric series converges as follows:

\[{S}_{n}=\sum _{i=1}^{n}{a}{r}^{i-1}=\cfrac{{a} (1 - {r}^{n})}{1 - r}\]

Now consider the behaviour of \({r}^{n}\) for \(-1

Let \(r=\cfrac{1}{2}\):

\begin{align*} n=1& : {r}^{n}={r}^{1}={\left(\cfrac{1}{2}\right)}^{1}=\cfrac{1}{2} \\ n=2& : {r}^{n}={r}^{2}={\left(\cfrac{1}{2}\right)}^{2}=\cfrac{1}{2} \cdot \cfrac{1}{2}=\cfrac{1}{4}<\cfrac{1}{2} \\ n=3& : {r}^{n}={r}^{3}={\left(\cfrac{1}{2}\right)}^{3}=\cfrac{1}{2} \cdot \cfrac{1}{2} \cdot \cfrac{1}{2}=\cfrac{1}{8} <\cfrac{1}{4} \end{align*}

Since \(r\) is in the range \(-1

Therefore,

\begin{align*}{S}_{n}& = \cfrac{{a}(1 - {r}^{n} )}{1 - r} \\\text{If } -1 < r < 1, \quad & \text{then } r^{n} arrow 0 \text{ as } n arrow \infty \\\therefore {S}_{\infty }& = \cfrac{{a}(1 - 0 )}{1 - r} \\ & = \cfrac{{a}}{1-r} \end{align*}

The sum of an infinite geometric series is given by the formula

\[\therefore {S}_{\infty }=\sum _{i=1}^{\infty }{a}{r}^{i-1}=\cfrac{{a}}{1-r} \qquad (-1where

\({a}\) is the first term of the series;
\(r\) is the constant ratio.

Alternative notation:

\[\underbrace{{S}_{n}}_{n arrow \infty} arrow \cfrac{a}{1 - r} \quad \text{ if }-1

In words: as the number of terms \((n)\) tends to infinity, the sum of a converging geometric series \((S_{n})\) tends to the value \(\cfrac{a}{1 - r}\).

Example

Question

Given the series \(18 + 6 + 2 + \cdots\). Find the sum to infinity if it exists.

Determine the value of \(r\)

We need to know the value of \(r\) to determine whether the series converges or diverges.

\begin{align*} \cfrac{T_{2}}{T_{1}} &= \cfrac{6}{18} \\ &= \cfrac{1}{3} \\ \cfrac{T_{3}}{T_{2}} &= \cfrac{2}{6} \\ &= \cfrac{1}{3} \\ \therefore r &= \cfrac{1}{3} \end{align*}

Since \(- 1 < r < 1\), we can conclude that this is a convergent geometric series.

Determine the sum to infinity

Write down the formula for the sum to infinity and substitute the known values:

\[a = 18; \qquad r = \cfrac{1}{3}\]

\begin{align*} S_{\infty} &= \cfrac{a}{1 - r} \\ &= \cfrac{18}{ 1 - \cfrac{1}{3}} \\ &= \cfrac{18}{\cfrac{2}{3}} \\ &= 18 \times \cfrac{3}{2} \\ &= 27 \end{align*}

As \(n\) tends to infinity, the sum of this series tends to \(\text{27}\); no matter how many terms are added together, the value of the sum will never be greater than \(\text{27}\).

Example

Question

Use two different methods to convert the recurring decimal \(0,\dot{5}\) to a proper fraction.

Convert the recurring decimal to a fraction using equations

\begin{align*} \text{Let } x &= \text{0.}\dot{\text{5}}\\ \therefore x &= \text{0.555} \ldots \ldots (1) \\ 10x &= \text{5.55} \ldots \ldots (2) \\ (2) - (1): \quad 9x &= 5 \\ \therefore x &= \cfrac{5}{9} \end{align*}

Convert the recurring decimal to a fraction using the sum to infinity

\begin{align*} \text{0.}\dot{\text{5}} &= \text{0.5} + \text{0.05} + \text{0.005} + \ldots \\ \text{or } \quad \text{0.}\dot{\text{5}} &= \cfrac{5}{10} + \cfrac{5}{100} + \cfrac{5}{1000} + \ldots \end{align*}

This is a geometric series with \(r = \text{0.1} = \cfrac{1}{10}\). And since \(- 1 < r < 1\), we can conclude that the series is convergent.

\begin{align*} S_{\infty} &= \cfrac{a}{1 - r} \\ &= \cfrac{\frac{5}{10}}{1 - \cfrac{1}{10}} \\ &= \dfrac{\cfrac{5}{10}}{\cfrac{9}{10}} \\ &= \cfrac{5}{9} \end{align*}

Example

Question

Determine the possible values of \(a\) and \(r\) if

\[\sum _{n=1}^{\infty}{ar^{n-1}} = 5\]

Write down the sum to infinity formula and substitute known values

\begin{align*} S_{\infty} &= \cfrac{a}{1 - r} \\ \therefore 5 &= \cfrac{a}{1 - r} \\ a &= 5(1 - r) \\ \therefore a &= 5 - 5r \\ \text{And } 5r &= 5 - a \\ \therefore r &= \cfrac{5 - a}{5} \end{align*}

Apply the condition for convergence to determine possible values of \(a\)

For a series to converge: \(- 1 < r < 1\)

\begin{align*} -1 & < r < 1 \\ -1 & < \cfrac{5 - a}{5} < 1 \\ -5 & < 5 - a < 5 \\ -10 & < - a < 0 \\ 0 & < a < 10 \end{align*}